A certain weight of pure `CaCO_(3)` is made to react completely with 200mL of a HCl solution to given 227mL of `CO_(2)` gas at STP. The normality of the HCl solution is :

Answer

To find the normality of the HCl solution that reacts with calcium carbonate (CaCO₃) to produce carbon dioxide (CO₂), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between calcium carbonate and hydrochloric acid can be written as:
\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \]
This equation shows that 1 mole of CaCO₃ reacts with 2 moles of HCl to produce 1 mole of CO₂.

2. **Determine Moles of CO₂ Produced:**
Given that 227 mL of CO₂ is produced at STP, we can calculate the moles of CO₂ using the molar volume of a gas at STP, which is 22.4 L (or 22400 mL):
\[ \text{Moles of CO}_2 = \frac{\text{Volume of CO}_2 \text{ (mL)}}{\text{Molar Volume (mL)}} = \frac{227 \text{ mL}}{22400 \text{ mL}} = 0.0101 \text{ moles} \]

3. **Calculate Moles of HCl:**
From the balanced equation, we see that 2 moles of HCl are required for every mole of CO₂ produced. Therefore, the moles of HCl can be calculated as:
\[ \text{Moles of HCl} = 2 \times \text{Moles of CO}_2 = 2 \times 0.0101 = 0.0202 \text{ moles} \]

4. **Calculate the Number of Gram Equivalents of HCl:**
The n-factor for HCl (which is a monoprotic acid) is 1. Thus, the number of gram equivalents of HCl is equal to the number of moles:
\[ \text{Gram Equivalents of HCl} = \text{Moles of HCl} \times \text{n-factor} = 0.0202 \times 1 = 0.0202 \text{ equivalents} \]

5. **Calculate Normality of HCl:**
Normality (N) is defined as the number of gram equivalents per liter of solution. The volume of HCl solution is given as 200 mL, which is 0.200 L. Therefore, we can calculate the normality as follows:
\[ \text{Normality} = \frac{\text{Gram Equivalents of HCl}}{\text{Volume of solution in L}} = \frac{0.0202}{0.200} = 0.101 \text{ N} \]

### Final Answer:
The normality of the HCl solution is approximately **0.1 N**.

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